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Chapter 6: Problem 71

A: $\begin{aligned} f(x) &=x^{4}-8 x^{3}+22 x^{2}-24 x+21 \\ f^{\prime}(x) &=4 x^{3}-24 x^{2}+44 x-24 \\ &=4\left(x^{3}-6 x^{2}+11 x-6\right) \\\ &=4(x-1)\left(x^{2}-5 x+6\right) \\ &=4(x-1)(x-2)(x-3) \end{aligned}$

Short Answer

Expert verified
Question: Verify if the given derivative is correct for the function \(f(x) = x^4 - 8x^3 + 22x^2 - 24x + 21\), where the derivative is given as \[f'(x) = 4x^3 - 24x^2 + 44x - 24 = 4(x-1)(x-2)(x-3)\] If necessary, make any corrections. Answer: The given derivative is correct.

Step by step solution

01

Apply the power rule

To find the derivative of \(f(x)\), we will use the power rule which states that if \(f(x)=x^n\), then \(f'(x)=nx^{n-1}\). Apply the power rule to each term in the function \(f(x)\): \[f'(x) = \frac{d}{dx}(x^4) - \frac{d}{dx}(8x^3) + \frac{d}{dx}(22x^2) - \frac{d}{dx}(24x) + \frac{d}{dx}(21)\]
02

Find the derivative of each term

Apply the power rule to each term and simplify: \[\begin{aligned} f'(x) &= 4x^3 - 24x^2 + 44x - 24 \end{aligned}\] Our result matches the given derivative: \[f'(x) = 4x^3 - 24x^2 + 44x - 24\]
03

Factor the derivative

Factor the correct derivative to match the given factorization: \[\begin{aligned} f'(x) &= 4x^3 - 24x^2 + 44x - 24 \\ &= 4(x^3 - 6x^2 + 11x - 6) \\ &= 4(x-1)(x-2)(x-3) \end{aligned}\] The factorization agrees with the given factorization, so the given derivative is correct. In conclusion, the derivative \(f'(x) = 4x^3 - 24x^2 + 44x - 24\) is correct and can be factored as \(f'(x) = 4(x-1)(x-2)(x-3)\).

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Most popular questions from this chapter

$\mathrm{f}^{\prime}\left(\mathrm{x}^{2}-4 \mathrm{x}+3\right)>0 \quad \forall \mathrm{x} \in(2,3)$ Now \(x^{2}-4 x+3=(x-3)(x-1)\) For \(x \in(2,3), x^{2}-4 x+3\) is increasing function from \((-1,0)\) Now, let \(g(x)=f(\sin x)\) \(g^{\prime}(x)=f^{\prime}(\sin x) \cos x\) Hence, \(\mathrm{C}\) is correct

\(f(x)=x^{3}+4 x^{2}+\lambda x+1\) \(f^{\prime}(x)=3 x^{2}+8 x+\lambda<0\) roots are \(-2 \& \frac{-2}{3}\) Product \(=+\frac{4}{3}\) \(\Rightarrow \lambda=+4\) Hence, A is correct

$\mathrm{f}^{\prime \prime}(\mathrm{x})-3 \mathrm{f}^{\prime}(\mathrm{x})>3, \mathrm{f}^{\prime}(0)=-1$ \(g(x)=f(x)+x\) \(g^{\prime}(x)=f^{\prime}(x)+1\) $\mathrm{g}^{\prime \prime}(\mathrm{x})=\mathrm{f}^{\prime \prime}(\mathrm{x})$ Now \(f^{\prime \prime}(x)>3\left(f^{\prime}(x)+1\right)\) $\frac{\mathrm{f}^{\prime \prime}(\mathrm{x})}{\mathrm{f}^{\prime}(\mathrm{x})+1}>3$ Integrating $=\mathrm{f}^{\prime}(\mathrm{x})+1>\mathrm{e}^{3 \mathrm{x}+\mathrm{c}}>0$ \(\Rightarrow g^{\prime}(x)>0\) Hence, B is correct

As \(f(x)\) is increasing function \(\Rightarrow 2 a^{2}+a+1>3 a^{2}-4 a+1\) \(a^{2}-5 a<0\) \(a(a-5)<0\) as \(x \in(0, \infty), \quad a \in\\{2,3,4\\}\) Hence, \(\mathrm{B}, \mathrm{C}, \mathrm{D}\) is correct

A) $\lim _{n \rightarrow \infty} \frac{1}{n} \sum \frac{1}{\sqrt{r / n}(3 \sqrt{r / n}+4)^{2}}=\int_{0}^{4} \frac{1}{\sqrt{x}(3 \sqrt{x}+4)^{2}} d x$ Put \(x=t^{2}\) $=2 \int_{0}^{2} \frac{\mathrm{dt}}{(3 \mathrm{t}+4)^{2}}=-\frac{2}{3}\left(\frac{1}{3 \mathrm{t}+4}\right)_{0}^{2}=-\frac{2}{3}\left[\frac{1}{10}-\frac{1}{4}\right]$ \(=\frac{1}{10}=\frac{\mathrm{p}}{\mathrm{q}}\) \(p+q=11\) B) \(f(x)=x^{3}+a x+2\) $\mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}+\mathrm{a}=0 \Rightarrow \mathrm{x}=\pm \alpha\( where \)\int=\sqrt{\frac{-\mathrm{a}}{\mathrm{f}}}$ \(f|\alpha| \cdot \leq f|-\alpha|>0 \Rightarrow a>-3\) Hence \(\mathrm{a}=-2,-1\) C) eqn of circle \(\Rightarrow x^{2}+y^{2}+2 g x+2 f y-1=0\) As it passes through \((1,1) \&(-2,1)\) $\Rightarrow 2 \mathrm{~g}+2 \mathrm{f}+1=0 \&$ \(-4 g+2 f+4=0\) Solving, \(\mathrm{g}=\frac{1}{2}, \mathrm{f}=-1\) \(f-c=0\) D) As \(x^{2}+y^{2}+z^{2}+2 x y z=-1\) \(\Rightarrow \sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\pi / 2\) Differentiate \(\frac{\mathrm{dx}}{\sqrt{1-\mathrm{x}^{2}}}+\frac{\mathrm{dy}}{\sqrt{1-\mathrm{y}^{2}}}+\frac{\mathrm{dz}}{\sqrt{1-\mathrm{z}^{2}}}=0\)
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